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Whitepunk [10]
3 years ago
7

What type of wave allows you to hear sounds?

Physics
2 answers:
DIA [1.3K]3 years ago
8 0
Electromagnetic transverse waves
Galina-37 [17]3 years ago
6 0
Electromagnetic transverse waves 
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If you know that the universe is estimated to be 14 billion years old . what is the age of the universe in seconds ???? (knowing
Svetlanka [38]

Answer:

<em>The estimated age of the universe is</em> 4.42*10^{17}\ seconds

Explanation:

<u>Time Units Conversion</u>

Sometimes we need to express magnitudes in different units, depending on the specific situation handled in the question or problem at hand. Magnitudes like length, time, mass, temperature, pressure, etc., can be expressed in several different units.

The most common units for time are years, months, days, hours, minutes, and seconds. There are 3,600 seconds in one hour and 24 hours in one day.

We are given the estimated age of the universe as 14 billion solar years. If one solar year is 365.25 days long, then the age, expressed in days is

14,000,000,000 * 365.25 days

Some quantities need to be expressed in scientific notation. This is the case since the numbers are too high

14,000,000,000 * 365.25 = 5.1135*10^{12}\ days

There are 24*3600=86400 seconds in one day, so

5.1135*10^{12}\ days=5.1135*10^{12} * 86400 =4.42*10^{17}\ seconds

The estimated age of the universe is 4.42*10^{17}\ seconds

6 0
3 years ago
Read 2 more answers
An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider compl
horsena [70]

Answer:

     A = 2,8333  s

Explanation:

El periodo es definido como el tiene que toma de dar una oscilación.

En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.

              T = t/n

calculemos

              A = 34,0/ 12,0

              A = 2,8333  s

6 0
2 years ago
You have 2 minutes to get to PE from science class before you get a tardy. If PE is 100m away and you walk at a speed of 1.1m/s
mart [117]

Answer:

D

Explanation:

4 0
3 years ago
Read 2 more answers
A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of
posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0
3 years ago
An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad
Anna11 [10]

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the  linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a)  where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

5 0
3 years ago
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